Chapter 7.1, Problem 52E

To determine

Solve the system graphically or algebraically. Explain your choice of method.

  $\{\begin{array}{c}y=x^3-2x^2+x-1\\ y=-x^2+3x-1\end{array}$

Answer to Problem 52E

The solution for the equations given is $x=2,y=1$ and $x=-1,y=-5$

Explanation of Solution

Given:

The given equations are $\{\begin{array}{c}y=x^3-2x^2+x-1\\ y=-x^2+3x-1\end{array}$

Concept Used:

The given equations are in two variables x and y. These equations can be solved using graphical method and algebraic method. This equations are simple so they can be solved algebraically.

By substituting one equation in the other we can solve the equations and find x and y values.

Calculation:

By equating both the equations we get x value.

  $\begin{array}{l}{x}^3-2x^2+x-1=-x^2+3x-1\\ x^3-x^2-2x=0\\ x\left(x^2-x-2\right)=0\\ x^2-2x+x-2=0\\ x\left(x-2\right)+1\left(x-2\right)=0\\ \left(x-2\right)\left(x+1\right)=0\\ x=2,-1\end{array}$

As it is a quadratic equation we get two values of x and by substituting them we get y values.

For $x=2$ , we get

  $\begin{array}{l}y=-x^2+3x-1\\ y=-{\left(2\right)}^2+3\left(2\right)-1\\ y=-4+6-1\\ y=1\end{array}$

For $x=-1$ , we get

  $\begin{array}{l}y=-x^2+3x-1\\ y=-{\left(-1\right)}^2+3\left(-1\right)-1\\ y=-1-3-1\\ y=-5\end{array}$

Conclusion:

The solution for the equations given is $x=2,y=1$ and $x=-1,y=-5$